A 4000 kg truck traveling at constant velocity and a braking force of 20000 N is suddenly applied. It takes 6 seconds for the truck to stop.

a) what is the deceleration of the truck?
b) what is the velocity of the truck before the brakes were applied
c) what distance did the truck travel from the time the brakes were applied to when it stopped.

F=ma

a=Fmax/m

if it takes six seconds...
Vf=Vi - F/m
solve for Vi

distance?
Vf^2=Vi^2-2ad solve for d.

To solve these questions, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = m * a). We can also use the kinematic equation, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

a) To find the deceleration of the truck, we can rearrange the equation F = m * a to solve for acceleration (a = F/m). Plugging in the values, we have a = 20000 N / 4000 kg = 5 m/s^2.

b) We know that the truck was traveling at a constant velocity before the brakes were applied. This means the acceleration was zero, and using the kinematic equation v = u + at, we can deduce that the final velocity (v) is equal to the initial velocity (u). Therefore, the velocity of the truck before the brakes were applied is also zero.

c) To find the distance the truck traveled from the time the brakes were applied to when it stopped, we can use another kinematic equation, s = ut + 0.5at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Since the initial velocity (u) is zero and the acceleration (a) is -5 m/s^2 (negative because the truck is decelerating), we have s = 0 + 0.5 * (-5 m/s^2) * (6 s)^2 = -90 meters (negative because the displacement is in the opposite direction of the truck's motion).

Therefore,
a) The deceleration of the truck is 5 m/s^2.
b) The velocity of the truck before the brakes were applied is zero.
c) The truck traveled a distance of 90 meters from the time the brakes were applied to when it stopped in the opposite direction of its motion.